I take it you meant that you first multiplied both sides by x- 3 so you had
[tex]\frac{x^2+ 2x+ 3}{x^2+ 9}= \frac{Ax+ B}{x^2+ 9}(x- 3)+ C[/tex]
Taking x= 3, this is [tex]\frac{9+ 6+ 3}{9+ 9}= 1= C[/tex].

then multiply out equation to find A and B

x2+2x+3 = C(X2+9) + (Ax+B)/(X-3)

Combine terms to give:

On the right side you mean? And (Ax+ B)(x- 3), not (Ax+ B)/(x- 3).

x2+9+Ax2-3Ax+Bx-3B

= x2(1+A)+ x(B-3A) + 9 -3B

if coefficients = 0 then x2 = 0 then A = -1

The way you have it set up the coefficients are not equal to 0- they are equal to 1, 2, and 3, the coefficients on the left side of the equation. A+ 1= 1 gives A= 0. 9- 3B= 3 gives B= 2.
With A= 0, B= 2, B- 3A= 2 automatically.

9-3B = 3 therefore B = 2

B-3A = 0 therefore A = 2/3 but I think this answer is wrong? I now have 2 answers for A?

B-3A = 0 therefore A = 2/3 but I think this answer is wrong? I now have 2 answers for A?

Any ideas please and thanks.

Do not make any assumptions about A,B and C: just write your function f(x) as
[tex] f(x) =\frac{x^2+2x+3}{(x^2+9)(x-3)} = \frac{Ax+B}{x^2+9}+\frac{C}{x-3} =
\frac{(A+C)x^2 +(B-3A)x + (9C-3B)}{(x^2+9)(x-3)} [/tex]
and so get three equations for the three parameters A,B,C. Solve them and see what you get.

The way you have it set up the coefficients are not equal to 0- they are equal to 1, 2, and 3, the coefficients on the left side of the equation. A+ 1= 1 gives A= 0. 9- 3B= 3 gives B= 2.
With A= 0, B= 2, B- 3A= 2 automatically.

How did you find the Coefficients as I thought you could automatically say that they are equal to Zero?

This is what I am failing to understand i.e. why choose these and not just Zero?

Thanks again as I realise this must be obvious to you.

OK, that is a good question. So assume that there exist ##A## and ##B## such that

$$x^2+2x+3=(1+A)x^2+(B−3A)x+(9−3B)~~~~(1)$$

are true all ##x##. In particular, it must be satisfied for ##x=0##. This gets us the equation

[tex]3 = 9-3B~~~~(2)[/tex]

But it must also be true for ##x=1##, which gets us

[tex]1^2 + 2\cdot 1 + 3 = (1+A) + (B-3A) + (9 - 3B)~~~~(3)[/tex]

And it must also be true for ##x=-1##, which gets us

[tex](-1)^2 + 2\cdot (-1) + 3 = (1+A) -(B-3A) +(9 - 3B)~~~~(4)[/tex]

Subtracting ##(2)## from ##(3)## and ##(4)## leaves us with the two equations

[tex]1 + 2 = (1+A) + (B-3A)~~~~(3^\prime)[/tex]

and

[tex]1 - 2 = (1+A) - (B-3A)~~~~(4^\prime)[/tex]

Adding up these two gets us

[tex]2 = 2(1+A)[/tex]

or just ##(1+A) = 1## which I call ##(5)##. Subtracting ##(5)## from ##(3^\prime)## then gets us

[tex]2 = B-3A[/tex]

So we finally see that we have the equations

[tex]1+A = 1,~2= B-3A,~3=9-3B[/tex]

Of course, I didn't do all that just to find these equations. What I said was simply that since

$$x^2+2x+3=(1+A)x^2+(B−3A)x+(9−3B)$$

must be true for all ##x##, then all the coefficients must equal. So the coefficients of ##x^2## must equal, which gets us ##1+A=1##. The coefficients of ##x## must equal, which gets us ##2 = B-3A##, and the constant terms must equal, so ##3 = 9-3B##.
But then you can ask why must the coefficients equal? Well, my above (long) arguments proves it. In general, whenever you are in the situation that

[tex]Ax^2 + Bx + C = \alpha x^2 + \beta x + \gamma[/tex]

you can apply my argument above and get that ##A=\alpha##, ##B=\beta## and ##C=\gamma##. So all coefficients equal.

Yes, but this is in the precalculus forum, so I didn't want to use the easier way of derivatives.

I assumed that. But since they post all kinds of calculus and beyond stuff in that forum anyway, and students don't usually encounter partial fractions for anything before calculus, I'm guessing the OP is in a calculus course and might appreciate the derivative method.